$h(t) = 3t+7$ $g(n) = 6n+3-h(n)$ $f(n) = 5n^{2}+h(n)$ $ h(g(-3)) = {?} $
First, let's solve for the value of the inner function, $g(-3)$ . Then we'll know what to plug into the outer function. $g(-3) = (6)(-3)+3-h(-3)$ To solve for the value of $g$ , we need to solve for the value of $h(-3)$ $h(-3) = (3)(-3)+7$ $h(-3) = -2$ That means $g(-3) = (6)(-3)+3-(-2)$ $g(-3) = -13$ Now we know that $g(-3) = -13$ . Let's solve for $h(g(-3))$ , which is $h(-13)$ $h(-13) = (3)(-13)+7$ $h(-13) = -32$